The proposed inequality $\|\log(RS)\|\le\|\log(R)+\log)S)\|$ does *not* hold for all $R,S\in\mathrm{SO}(3)$ for which $\log(RS)$, $\log(R)$, and $\log(S)$ are defined.  (In what follows, I'm taking the Frobenius norm to be $\|B\|^2 = \tfrac12\,\mathrm{tr}(B^TB)$, but this normalization clearly does not affect the argument.)

First, of all, notice that the function $B\mapsto \|\log(B)\|$ is actually a continuous function on $\mathrm{SO}(3)$, because of the easily checked identity
$$
\cos\bigl(\|\log(B)\|\bigr) = \tfrac12\bigl(\mathrm{tr}(B)-1\bigr)
$$
and the fact that $\cos^{-1}:[-1,1]\to[0,\pi]$ is a continuous, strictly decreasing function on $[-1,1]$.  If the desired inequality actually held, then we would have
$$
\mathrm{tr}(RS) 
\ge 1+ 2\,\cos\bigl(\|\log(R)+\log(S)\|\bigr).
$$
Now consider the two matrices
$$
R = \exp\begin{pmatrix}0&-x&0\\x&0&0\\0&0&0\end{pmatrix}
\quad\text{and}\quad
S = \exp\begin{pmatrix}0&0&-y\\0&0&0\\y&0&0\end{pmatrix},
$$
where $0\le x,y < \pi$ (so that $R$, $S$, and $RS$ remain in the domain of the $\log$ function).  The above inequality would imply
$$
\cos x\cos y+\cos x + \cos y \ge 1 + 2\,\cos\left(\sqrt{x^2+y^2}\right).
$$
However, this inequality does not hold:  Take $x=y$ in the above expressions. Then, indeed, the inequality holds for small $x\ge0$, but there is a value
$x=y\approx 2.405181<\pi$ where both sides of this last inequality become equal and then the inequality reverses in the range $2.405182\le x=y <\pi$. 

Thus, the original proposed inequality does not hold on $\mathrm{SO}(3)$.

**Remark:**  The situation is different for the unit quaternions (which can also be thought of as $\mathrm{SU}(2)$, but I won't.)
There, we do indeed have $|\log(pq)|\le |\log(p) + \log(q)|$
for unit quaternions $p$ and $q$ different from $-1$, as long as we make the convention that $|\log(-1)| = \pi$, even though $\log(-1)$ cannot be defined continuously.

Consider the quaternions, i.e., expressions of the form $q = x_0 + x_1\,i  + x_2\,j + x_3\,k$, where the $x_i$ are real numbers and $i$, $j$, and $k$ satisfy the usual relations $i^2=j^2=k^2=-1$ and $ij-k=jk-i=ki-j=ji+k=kj+i=ji+k=0$.  The *real part* of $q$ is $x_0$ and the *imaginary part* of $q$ is $x_1\,i  + x_2\,j + x_3\,k$. As usual, set $|q| = (x_0^2+x_1^2+x_2^2+x_3^2)^{1/2}$.

If $u$ is a unit imaginary quaternion then $u^2=-1$, so that, when $a$ is real, we have $\exp(a\,u) = \cos a + \sin a \,u$.  Every unit quaternion $q$ other than $-1$ is of the form $\exp(x)$ for some *unique* purely imaginary quaterion $x$ with $|x|<\pi$. Then setting $x := \log q $ defines $\log$ as a smooth function on the unit quaternions minus the single element $-1$.  Note that $|\log q|) = |x| = |a|$, so $\cos\bigl(|\log q|\bigr) = \cos a $, which is the real part of $q$.

Suppose that $q = \cos a + \sin a \,u$
while $p=\cos b + \sin b \,v$ where $0\le a,b<\pi$ and $u$ and $v$ are unit quaternions, with $u\cdot v = \cos c$ for some $c$.  Note that the real part of $uv$ is $-\cos c$.  Thus, we have the product expansion 
$$
qp = (\cos a + \sin a \,u)(\cos b + \sin b \,v) 
= (\cos a\cos b-\sin a\sin b\cos c) + X
$$
where $X$ is imaginary. Thus
$$
\cos\bigl(|\log(qp)|\bigr) = (\cos a\cos b-\sin a\sin b\cos c).
$$
Meanwhile 
$$
|\log q + \log p| = |au+bv| = \sqrt{a^2+2ab\cos c+b^2}.
$$
Thus, the inequality $|\log(qp)|\le |\log q + \log p|$ is equivalent to
$$
(\cos a\cos b-\sin a\sin b\cos c)
\ge \cos\bigl(\sqrt{a^2+2ab\cos c+b^2}\bigr),
$$
and this inequality does indeed hold for all $a$ and $b$ 
with $|a|,|b|<\pi$ and all $c$.  (In fact, by continuity, it holds for $|a|,|b|\le \pi$ and all $c$.  Note that equality holds in this range of the variables only when $a=0$, $b=0$ or $\cos c = \pm 1$.)

(Note that $qp=-1$ would imply that $p=-\bar q$, so $\cos(a)+\cos(b)=0$ and $\sin a\,u=\sin b\,v$.  But then $|\sin a|=|\sin b|$.  This forces $a+b = \pi$ (since we can assume that $0\le a,b <\pi$) and $u=v$, i.e., $c=0$.  Thus, the inequality holds even when $\log (qp)$ does not exist.)